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Dennis_Churaev [7]
2 years ago
10

Which choices are solutions to the following equation? Check all that apply. X^2-10x-5=9

Mathematics
1 answer:
Afina-wow [57]2 years ago
5 0
Solve using quadratic formula!
Answers:
5 + √39
5 - √39
You might be interested in
Solve this to help me
Mrac [35]

Answer:

D

Step-by-step explanation:

Hope I helped

6 0
2 years ago
The average age of 4 children is 15 yrs . if the ages of their parents are added , their average age is 25 .find the age of thei
WITCHER [35]

The age of the father is 57 years old

<h3>Calculating mean of a population</h3>

Total age = Average age x Population

Average of the four children = 15

Population = 4

Total age of the four children = 15 x 4

Total age of the four children = 60 years

If the parents are added:

Population = 6

Average age = 25

Total age = 25 x 6

Total age = 150 years

Total age of the parents = 150 - 60

Total age of the parents = 110 years

The father is 4 years older than his wife

Let the age of the father be x

The age of the wife = x - 4

x + x - 4 = 110

2x = 114

x = 114/2

x = 57

Therefore, the age of the father is 57 years old

Learn more on The mean of numbers here: brainly.com/question/20118982

#SPJ1

6 0
2 years ago
What are the zeros of f(x) = x(x − 9)?
ollegr [7]

Answer:

x = 0 , x = 9

Step-by-step explanation:

to find the zeros let f(x) = 0 , that is

x(x - 9) = 0

equate each factor to zero and solve for x

x = 0

x - 9 = 0 ⇒ x = 9

6 0
2 years ago
Read 2 more answers
Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo
Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0
3 years ago
If Colorado Springs, Colorado, has 1.4 times more days of sunshine than Boston, Massachusetts, how many days of sunshine does ea
Rom4ik [11]
Days in boston=x
days in colorado=1.4x
boston+colorado=2.4x
475=2.4x
475/2.4=2.4x/2.4
198=x
198*1.4=277
days in boston=198
days in colorado=277
198+277=475


5 0
3 years ago
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